1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.
2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.
Answer:
a.18.5 m/s
b.1.98 s
Explanation:
We are given that
![\theta=35^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D35%5E%7B%5Ccirc%7D)
a.Let
be the initial velocity of the ball.
Distance,x=30 m
Height,h=1.8 m
![v_x=v_0cos\theta=v_0cos35](https://tex.z-dn.net/?f=v_x%3Dv_0cos%5Ctheta%3Dv_0cos35)
![v_y=v_0sin\theta=v_0sin35](https://tex.z-dn.net/?f=v_y%3Dv_0sin%5Ctheta%3Dv_0sin35)
![x=v_0cos\theta\times t=v_0cos35\times t](https://tex.z-dn.net/?f=x%3Dv_0cos%5Ctheta%5Ctimes%20t%3Dv_0cos35%5Ctimes%20t)
![t=\frac{30}{v_0cos35}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B30%7D%7Bv_0cos35%7D)
![h=v_yt-\frac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dv_yt-%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substitute the values
![1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2](https://tex.z-dn.net/?f=1.8%3Dv_0sin35%5Cfrac%7B30%7D%7Bv_0cos35%7D-%5Cfrac%7B1%7D%7B2%7D%289.8%29%28%5Cfrac%7B30%7D%7Bv_0cso35%7D%29%5E2)
![1.8=30tan35-\frac{6574.6}{v^2_0}](https://tex.z-dn.net/?f=1.8%3D30tan35-%5Cfrac%7B6574.6%7D%7Bv%5E2_0%7D)
![\frac{6574.6}{v^2_0}=21-1.8=19.2](https://tex.z-dn.net/?f=%5Cfrac%7B6574.6%7D%7Bv%5E2_0%7D%3D21-1.8%3D19.2)
![v^2_0=\frac{6574.6}{19.2}](https://tex.z-dn.net/?f=v%5E2_0%3D%5Cfrac%7B6574.6%7D%7B19.2%7D)
![v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s](https://tex.z-dn.net/?f=v_0%3D%5Csqrt%7B%5Cfrac%7B6574.6%7D%7B19.2%7D%7D%3D18.5%20m%2Fs)
Initial velocity of the ball=18.5 m/s
b.Substitute the value then we get
![t=\frac{30}{18.5cos35}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B30%7D%7B18.5cos35%7D)
t=1.98 s
Hence, the time for the ball to reach the target=1.98 s
Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.
Answer:
yes 20 characters or more
Explanation: